Find $\lim_{x\to\scriptsize\dfrac{\pi}{2}}\dfrac{\cot^2(x)}{1-\sin(x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $-\dfrac{\pi}{2}$ (Choice C) C $2$ (Choice D) D The limit doesn't exist
Answer: Substituting $x=\dfrac{\pi}{2}$ into $\dfrac{\cot^2(x)}{1-\sin(x)}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since our expression includes trigonometric functions, let's try to re-write it using factorization and trigonometric identities. Since we have $\cos^2(x)$ in our expression, because $\cot^2(x)=\dfrac{\cos^2(x)}{\sin^2({x})}$, let's rewrite it using the Pythagorean identity, $\sin^2(x)+\cos^2(x)=1$ : $\begin{aligned} &\phantom{=}\dfrac{\cot^2(x)}{1-\sin(x)} \\\\ &=\dfrac{\cos^2({x})}{\sin^2(x)[1-\sin(x)]} \gray{\text{Definition of cotangent}} \\\\ &=\dfrac{(1-\sin^2(x))}{\sin^2(x)[1-\sin(x)]} \gray{\text{The Pythagorean identity}} \\\\ &=\dfrac{(1-\sin(x))(1+\sin(x))}{\sin^2(x)[1-\sin(x)]} \gray{\text{Diff. of squares}} \\\\ &=\dfrac{\cancel{(1-\sin(x))}(1+\sin(x))}{\sin^2(x)[\cancel{1-\sin(x)}]} \gray{\text{Cancel common factors}} \\\\ &=\dfrac{1+\sin(x)}{\sin^2(x)}\text{, for }x\neq \{...,- \dfrac{7\pi}{2}, - \dfrac{3\pi}{2}, \dfrac{\pi}{2}, \dfrac{5\pi}{2}, \dfrac{9\pi}{2},...\} \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $(2k+1)\dfrac{\pi}{2}$ for any integer $k$, and specifically $\dfrac{\pi}{2}$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{\cot^2(x)}{1-\sin(x)}=\dfrac{1+\sin(x)}{\sin^2(x)}$ for all $x$ -values in the interval $(-\pi,\pi)$ except for $x=\dfrac{\pi}{2}$. Therefore $\lim_{x\to \scriptsize\dfrac{\pi}{2}}\dfrac{\cot^2(x)}{1-\sin(x)}=\lim_{x\to \scriptsize\dfrac{\pi}{2}}\dfrac{1+\sin(x)}{\sin^2(x)}=2$ (The last limit was found using direct substitution.) In conclusion, $\lim_{x\to\scriptsize\dfrac{\pi}{2}}\dfrac{\cot^2(x)}{1-\sin(x)}=2$.